I want to evaluate an integral which is essentially a sum of the form $\lim_{a\to 0^+}\int_{-\infty}^\infty \frac{e^{ikx}}{(x+ia)^m} dx$, where $k\in\mathbb R$ and $m\in\mathbb N$. By the residue theorem, we can easily evaluate the integral $$\int_{-\infty}^\infty \frac{e^{ikx}}{(x+ia)^m} dx = -\frac{2\pi i}{(m-1)!} (ik)^{m-1} e^{ka} \theta(-k), \tag{*}$$ where $\theta$ is the Heaviside function. In particular, letting $a\to 0^+$, we simplify $$\lim_{a\to 0^+}\int_{-\infty}^\infty \frac{e^{ikx}}{(x+ia)^m} dx = -\frac{2\pi i}{(m-1)!} (ik)^{m-1} \theta(-k).$$
Now, I have a somewhat complicated expression, which is obtained from other Mathematica calculation:
$Assumptions = {a > 0, k1 > 0, k2 > 0, k3 > 0, k1 + k2 - k3 > 0}
exp = 1/(x3 + I a)^3 E^(-I (k1 + k2 -
k3) x3) (-3 I (-1 + E^(I k1 x3)) (-1 + E^(I k2 x3)) k2 +
3 k1 (I + I E^(I (k1 + k2) x3) + E^(I k1 x3) (-I + k2 x3) -
E^(I k2 x3) (I + k2 x3)))
Question: I want to evaluate
Integrate[exp, {x3, -Infinity, Infinity}]/.{a->0}
Although exp
is complicated, it is a sum of the form $(*)$, so it should be calculable. However, Mathematica cannot evaluate this in a reasonable time.
I think I have to somehow "teach" Mathematica to use $(*)$. How can I do this?
Integrate[Simplify@#, {x3, -Infinity, Infinity}, Assumptions -> {a > 0, k1 > 0, k2 > 0, k3 > 0, k1 + k2 - k3 > 0}] & /@ Expand[exp]
$\endgroup$FourierTransform
, though. $\endgroup$